For an equilibrium reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8 xx 10^(-2)` and `1.2 xx 10^(-2) mol//L` respectively. The value of `K_(c)` for the reaction is

Options

A. `3 xx 10^(-3)mol//L`
B. `3.3 xx 10^(-3) xx 10^(-3)mol//L`
C. `3 xx 10^(-1) mol//L`
D. `3.3 xx 10^(-1)mol//L`

Answer

A

Explanation

According to law of active mass
`K_(C) = ([NO_(2)]^(2))/([N_(2)O_(4)]) = ([1.2 xx 10^(-2)]^(2))/(4.8xx 10^(-2))`
`= 0.3xx 10^(-2) = 3xx 10^(-3) mol//L`